Let $(X,d)$ be a locally compact metric space. Then for each $x \in X$ $\exists$ $\epsilon_x > 0$ such that $B[x;\epsilon_x] = \{y \in X : d(x,y) \leq \epsilon_x \}$ is compact.

How do I proceed to prove it? Please help me in this regard.

Thank you very much.


Pretty simple!

Let us choose $x \in X$ arbitrarily. Since $X$ is locally compact so $\exists$ a compact neighbourhood $C_x$ of $x$ in $X$ i.e. $\exists$ $\epsilon >0$ such that $x \in B(x;\epsilon) \subset C_x$. Consider $0 < \epsilon_x < \epsilon$ then clearly $B[x;\epsilon_x] \subset B(x;\epsilon) \subset C_x$. Now $B[x;\epsilon]$ (being a closed subset of a compact set $C_x$) is compact. Which proves your claim.

  • $\begingroup$ Actually it is a necessary and sufficient condition for locally compactness. $\endgroup$ – Dbchatto67 Aug 19 '18 at 6:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.